Question

differential equations in the form of F(x,y,p), F(y/x,p) or differential equations containing p(p=dy/dx)

1. (y-xp)^2=b^2+a^2 p^2
2. y^2 p^2+4y^2-(x+yp)^2=0

Answer 1

ANSWER on Question #79724 – Math – Differential Equations

QUESTION

differential equations in the form of $F(x,y,p)$, $F(y/x,p)$ or differential equations containing $p(p = dy/dx)$

$$\begin{array}{l} 1. (y - x p) ^ {2} = b ^ {2} + a ^ {2} p ^ {2} \\ 2. y ^ {2} p ^ {2} + 4 y ^ {2} - (x + y p) ^ {2} = 0 \\ \end{array}$$

SOLUTION

1. Solve equation $y^{2}p^{2} + 4y^{2} - (x + yp)^{2} = 0$

$$y ^ {2} p ^ {2} + 4 y ^ {2} - (x + y p) ^ {2} = 0 \rightarrow y ^ {2} p ^ {2} + 4 y ^ {2} - x ^ {2} - 2 x y p - y ^ {2} p ^ {2} = 0 \rightarrow 2 x y p = 4 y ^ {2} - x ^ {2} \rightarrow$$ $$2 x y p = 4 y ^ {2} - x ^ {2} \mid \div (x) \rightarrow 2 y p = \frac {4 y ^ {2}}{x} - x \rightarrow \boxed {2 y \cdot \frac {d y}{d x} = \frac {4 y ^ {2}}{x} - x \text{ is Bernoulli differential equation}}$$

(More information: https://en.wikipedia.org/wiki/Bernoulli_differential_equation)

In our case,

2 y \cdot \frac {d y}{d x} = \frac {4 y ^ {2}}{x} - x \rightarrow \left[\begin{array}{c}\text{We introduce the substitution}\\\text{$y^2 = u \rightarrow 2y \cdot \frac{dy}{dx} = \frac{du}{dx}}\end{array}\right]\rightarrow \boxed{\frac {d u}{d x} = \frac {4 u}{x} - x \text{ is inhomogeneous equation}}

1 STEP: Solve the homogeneous equation

$$\frac {d u}{d x} = \frac {4 u}{x} \rightarrow \frac {d u}{u} = \frac {4 d x}{x} \rightarrow \ln | u | = 4 \ln | x | + \ln | C | \rightarrow u = C \cdot x ^ {4}$$

2 STEP: Solve the inhomogeneous equation.

We apply the method of variation of the constant.

$$C = C (x) \rightarrow u = C (x) \cdot x ^ {4} \rightarrow \frac {d u}{d x} = \frac {d C}{d x} \cdot x ^ {4} + 4 C x ^ {3}$$

Then,

$$\begin{array}{l} \frac {d u}{d x} = \frac {4 u}{x} - x \rightarrow \frac {d C}{d x} \cdot x ^ {4} + 4 C x ^ {3} = \frac {4 C \cdot x ^ {4}}{x} - x \rightarrow \frac {d C}{d x} \cdot x ^ {4} + 4 C x ^ {3} = 4 C x ^ {3} - x \rightarrow \\ \frac {d C}{d x} \cdot x ^ {4} = - x \rightarrow d C = - \frac {d x}{x ^ {3}} \rightarrow \boxed {C (x) = \frac {1}{2 x ^ {2}} + C} \\ \end{array}$$

Conclusion,

$$u = C(x) \cdot x^4 = x^4 \cdot \left(\frac{1}{2x^2} + C\right) \rightarrow \boxed{u = \frac{x^2}{2} + C \cdot x^4}$$

Then,

$$\left\{ \begin{array}{c} u = \frac{x^2}{2} + C \cdot x^4 \rightarrow y^2 = \frac{x^2}{2} + C \cdot x^4 \rightarrow \boxed{y = \pm \sqrt{\frac{x^2}{2} + C \cdot x^4}} \end{array} \right.$$

2. Solve equation $(y - xp)^2 = b^2 + a^2 p^2$

1. STEP: Let's look at this equation as an algebraic one. We express the variable $y$ in terms of $x, p, a, b$.

$$(y - xp)^2 = b^2 + a^2 p^2 \rightarrow y - xp = \pm \sqrt{b^2 + a^2 p^2} \rightarrow \boxed{y = xp \pm \sqrt{b^2 + a^2 p^2}}$$

2. STEP: We differentiate the resulting equation, remembering that $p = \frac{dy}{dx}$

$$\left(Note: p' = \frac{dp}{dx}\right)$$ $$\frac{d}{dx} \times \left| y = xp \pm \sqrt{b^2 + a^2 p^2} \rightarrow \frac{dy}{dx} \right. = p = \frac{d}{dx} \left(xp \pm \sqrt{b^2 + a^2 p^2}\right) = p + xp' \pm \frac{2a^2 p \cdot p'}{2\sqrt{b^2 + a^2 p^2}} \rightarrow$$ $$p = p + xp' \pm \frac{a^2 p \cdot p'}{\sqrt{b^2 + a^2 p^2}} \rightarrow xp' \pm \frac{a^2 p \cdot p'}{\sqrt{b^2 + a^2 p^2}} = 0 \rightarrow p' \cdot \left(x \pm \frac{a^2 p}{\sqrt{b^2 + a^2 p^2}}\right) = 0 \rightarrow$$ $$\boxed{ \begin{array}{c} 1 \text{ case: } p' = 0 \\ 2 \text{ case: } x \pm \frac{a^2 p}{\sqrt{b^2 + a^2 p^2}} = 0 \end{array} }$$

1. case: $p' = 0$

$$p' = 0 \rightarrow p = Const = C_1 \rightarrow \left(p = \frac{dy}{dx}\right) \rightarrow \frac{dy}{dx} = C_1 \rightarrow dy = C_1 dx \rightarrow \boxed{y = C_1 x + C_2}$$

It remains to determine the constants $C_1$ and $C_2$. We substitute the solution found in the initial equation:

$$\left\{ \begin{array}{c} (y - xp)^2 = b^2 + a^2 p^2 \\ p = C_1 \\ y = C_1 x + C_2 \end{array} \right. \rightarrow (C_1 x + C_2 - C_1 x)^2 = b^2 + a^2 C_1^2 \rightarrow C_2^2 = b^2 + a^2 C_1^2 \rightarrow$$ $$\left\{ \begin{array}{c} C _ {1} = c \\ C _ {2} = \pm \sqrt {b ^ {2} + a ^ {2} c ^ {2}} \to y = c x \pm \sqrt {b ^ {2} + a ^ {2} c ^ {2}} \end{array} \right.$$

2 case:

$$x \pm \frac {a ^ {2} p}{\sqrt {b ^ {2} + a ^ {2} p ^ {2}}} = 0 \rightarrow x = \mp \frac {a ^ {2} p}{\sqrt {b ^ {2} + a ^ {2} p ^ {2}}} \Bigg | \times \sqrt {b ^ {2} + a ^ {2} p ^ {2}} \rightarrow x \sqrt {b ^ {2} + a ^ {2} p ^ {2}} = \mp a ^ {2} p \rightarrow$$ $$\left(x \sqrt {b ^ {2} + a ^ {2} p ^ {2}}\right) ^ {2} = (\mp a ^ {2} p) ^ {2} \rightarrow x ^ {2} (b ^ {2} + a ^ {2} p ^ {2}) = a ^ {4} p ^ {2} \rightarrow x ^ {2} b ^ {2} + a ^ {2} x ^ {2} p ^ {2} = a ^ {4} p ^ {2} \rightarrow$$ $$x ^ {2} b ^ {2} = a ^ {4} p ^ {2} - a ^ {2} x ^ {2} p ^ {2} \rightarrow x ^ {2} b ^ {2} = a ^ {2} (a ^ {2} - x ^ {2}) p ^ {2} | \div a ^ {2} (a ^ {2} - x ^ {2}) \rightarrow p ^ {2} = \frac {x ^ {2} b ^ {2}}{a ^ {2} (a ^ {2} - x ^ {2})} \rightarrow$$ $$\sqrt {p ^ {2}} = \sqrt {\frac {x ^ {2} b ^ {2}}{a ^ {2} (a ^ {2} - x ^ {2})}} \rightarrow p = \pm \frac {b}{a} \cdot \frac {x}{\sqrt {a ^ {2} - x ^ {2}}} \rightarrow \left(p = \frac {d y}{d x}\right)\rightarrow \frac {d y}{d x} = \pm \frac {b}{a} \cdot \frac {x}{\sqrt {a ^ {2} - x ^ {2}}} \rightarrow$$ $$d y = \pm \frac {b}{a} \cdot \frac {x \cdot d x}{\sqrt {a ^ {2} - x ^ {2}}} \rightarrow y = \pm \frac {b}{a} \cdot \int \frac {x d x}{\sqrt {a ^ {2} - x ^ {2}}} = \left[\begin{array}{c}a ^ {2} - x ^ {2} = t\\- 2 x d x = d t \rightarrow x d x = - \frac {d t}{2}\end{array}\right]\rightarrow$$ $$y = \pm \frac {b}{a} \cdot \int \frac {- \frac {d t}{2}}{\sqrt {t}} \rightarrow y = \mp \frac {b}{a} \cdot \int \frac {d t}{2 \sqrt {t}} \rightarrow y = \mp \frac {b}{a} \cdot \sqrt {t} + C o n s t \rightarrow \boxed {y = \mp \frac {b}{a} \cdot \sqrt {a ^ {2} - x ^ {2}} + C o n s t}$$

Conclusion,

$$\left\{ \begin{array}{c} y = c x \pm \sqrt {b ^ {2} + a ^ {2} c ^ {2}} \\ \text{or} \\ y = \mp \frac {b}{a} \cdot \sqrt {a ^ {2} - x ^ {2}} + \text{Const} \end{array} \right.$$

ANSWER:

1. $\left\{ \begin{array}{c}y = cx\pm \sqrt{b^2 + a^2c^2}\\ \text{or}\\ y = \mp \frac{b}{a}\cdot \sqrt{a^2 - x^2} +\text{Const} \end{array} \right.$

2. $y = \pm \sqrt{\frac{x^2}{2} + C\cdot x^4}$

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